Question

Tìm x để \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) ≤ \(\dfrac{1}{2}\)

Answer 1

Đk: \(x\ge0\)

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\le\dfrac{1}{2}\)

\(\Leftrightarrow1-\dfrac{2}{\sqrt{x}+1}\le\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\ge\dfrac{1}{2}\) \(\Leftrightarrow\sqrt{x}+1\le4\Leftrightarrow\sqrt{x}\le3\Leftrightarrow x\le9\)

Vậy BPT có nghiệm \(x\in\left[0;9\right]\)
Để gặp em nguyện F5

Answer 2

=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< =0\)

=>\(\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< =0\)

=>căn x-3<=0

=>x<=9 và x>=0